Mixing
Another variant / corollary of Langleys adventitious angles triangle problem
$begingroup$
I recently came across an elegant simple method on Youtube to solve the original Langley's problem using basic geometry principles. Worlds Hardest Easy Geometry problem
I have also gone through related (but not same) stack exchange postsworlds-hardest-easy-geometry-problem and
solve-for-x-in-the-80-circ-80-circ-20-circ-triangle.
Using that I was also trying to generalize the problem in a slightly different way. I assumed only the apex angle and the original triangle $ABC$ being Isosceles triangle along with some additional constraints detailed below. My objective is to find $x$ as a function of the apex angle $z$ only. The equations for the different angles are listed below.Langley's triangle with an added construction
With reference to my diagram:
$BG$ is constructed so that $CBG$ equals apex angle $z$.
$BF = BG$ (additional constraint I have added. This holds good for the original Langley). Basically $BC$, $BG$ and $BF$ are on an arc with the same radius.
$GF = GE$ (additional constraint I have added which again holds good for the original Langley).
With this and the set of Isosceles triangles we can derive (pardon the formatting, I am a newbie yet to absorb all the formatting rules)
$z + a + b = d + e$
$a + b = 90 - 3z /2$
$d = 45^{circ} + z/4$
$e = 45^{circ} - 3z /4$
$f = 45^{circ} + 3z /4$
$i = 45^{circ} - z / 4$
$j = 67.5^{circ} + z / 8$
$l = 90^{circ} - z / 2 - b$
$x = 5z / 8 + b - 22.5^{circ} = 67.5^{circ} - 7z/8 -a$
I have verified that if I use the original Langley's apex angle $z = 20^{circ}$ and $a = 20^{circ}$ (or $b = 40^{circ}$) I get the correct answer as $x = 30^{circ}$.
So except for $l$ and $x$ all other angles are determined solely by the apex angle $z$. By construction it appears that these two should also be uniquely determined once the apex angle is fixed.
So my question is that:
Is there a way to derive angle $x$ ($BEF$) as a function angle $z$ ($BAC$) alone?
Alternately can angle $a$ or angle $b$ derived as a function of the apex angle $z$?
triangle angle
edited Jan 17 at 10:05
idriskameni
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
$endgroup$
add a comment |
$begingroup$
I recently came across an elegant simple method on Youtube to solve the original Langley's problem using basic geometry principles. Worlds Hardest Easy Geometry problem
I have also gone through related (but not same) stack exchange postsworlds-hardest-easy-geometry-problem and
solve-for-x-in-the-80-circ-80-circ-20-circ-triangle.
Using that I was also trying to generalize the problem in a slightly different way. I assumed only the apex angle and the original triangle $ABC$ being Isosceles triangle along with some additional constraints detailed below. My objective is to find $x$ as a function of the apex angle $z$ only. The equations for the different angles are listed below.Langley's triangle with an added construction
With reference to my diagram:
$BG$ is constructed so that $CBG$ equals apex angle $z$.
$BF = BG$ (additional constraint I have added. This holds good for the original Langley). Basically $BC$, $BG$ and $BF$ are on an arc with the same radius.
$GF = GE$ (additional constraint I have added which again holds good for the original Langley).
With this and the set of Isosceles triangles we can derive (pardon the formatting, I am a newbie yet to absorb all the formatting rules)
$z + a + b = d + e$
$a + b = 90 - 3z /2$
$d = 45^{circ} + z/4$
$e = 45^{circ} - 3z /4$
$f = 45^{circ} + 3z /4$
$i = 45^{circ} - z / 4$
$j = 67.5^{circ} + z / 8$
$l = 90^{circ} - z / 2 - b$
$x = 5z / 8 + b - 22.5^{circ} = 67.5^{circ} - 7z/8 -a$
I have verified that if I use the original Langley's apex angle $z = 20^{circ}$ and $a = 20^{circ}$ (or $b = 40^{circ}$) I get the correct answer as $x = 30^{circ}$.
So except for $l$ and $x$ all other angles are determined solely by the apex angle $z$. By construction it appears that these two should also be uniquely determined once the apex angle is fixed.
So my question is that:
Is there a way to derive angle $x$ ($BEF$) as a function angle $z$ ($BAC$) alone?
Alternately can angle $a$ or angle $b$ derived as a function of the apex angle $z$?
triangle angle
edited Jan 17 at 10:05
idriskameni
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
$endgroup$
$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40
add a comment |
$begingroup$
I recently came across an elegant simple method on Youtube to solve the original Langley's problem using basic geometry principles. Worlds Hardest Easy Geometry problem
I have also gone through related (but not same) stack exchange postsworlds-hardest-easy-geometry-problem and
solve-for-x-in-the-80-circ-80-circ-20-circ-triangle.
Using that I was also trying to generalize the problem in a slightly different way. I assumed only the apex angle and the original triangle $ABC$ being Isosceles triangle along with some additional constraints detailed below. My objective is to find $x$ as a function of the apex angle $z$ only. The equations for the different angles are listed below.Langley's triangle with an added construction
With reference to my diagram:
$BG$ is constructed so that $CBG$ equals apex angle $z$.
$BF = BG$ (additional constraint I have added. This holds good for the original Langley). Basically $BC$, $BG$ and $BF$ are on an arc with the same radius.
$GF = GE$ (additional constraint I have added which again holds good for the original Langley).
With this and the set of Isosceles triangles we can derive (pardon the formatting, I am a newbie yet to absorb all the formatting rules)
$z + a + b = d + e$
$a + b = 90 - 3z /2$
$d = 45^{circ} + z/4$
$e = 45^{circ} - 3z /4$
$f = 45^{circ} + 3z /4$
$i = 45^{circ} - z / 4$
$j = 67.5^{circ} + z / 8$
$l = 90^{circ} - z / 2 - b$
$x = 5z / 8 + b - 22.5^{circ} = 67.5^{circ} - 7z/8 -a$
I have verified that if I use the original Langley's apex angle $z = 20^{circ}$ and $a = 20^{circ}$ (or $b = 40^{circ}$) I get the correct answer as $x = 30^{circ}$.
So except for $l$ and $x$ all other angles are determined solely by the apex angle $z$. By construction it appears that these two should also be uniquely determined once the apex angle is fixed.
So my question is that:
Is there a way to derive angle $x$ ($BEF$) as a function angle $z$ ($BAC$) alone?
Alternately can angle $a$ or angle $b$ derived as a function of the apex angle $z$?
triangle angle
edited Jan 17 at 10:05
idriskameni
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
$endgroup$
I recently came across an elegant simple method on Youtube to solve the original Langley's problem using basic geometry principles. Worlds Hardest Easy Geometry problem
I have also gone through related (but not same) stack exchange postsworlds-hardest-easy-geometry-problem and
solve-for-x-in-the-80-circ-80-circ-20-circ-triangle.
Using that I was also trying to generalize the problem in a slightly different way. I assumed only the apex angle and the original triangle $ABC$ being Isosceles triangle along with some additional constraints detailed below. My objective is to find $x$ as a function of the apex angle $z$ only. The equations for the different angles are listed below.Langley's triangle with an added construction
With reference to my diagram:
$BG$ is constructed so that $CBG$ equals apex angle $z$.
$BF = BG$ (additional constraint I have added. This holds good for the original Langley). Basically $BC$, $BG$ and $BF$ are on an arc with the same radius.
$GF = GE$ (additional constraint I have added which again holds good for the original Langley).
With this and the set of Isosceles triangles we can derive (pardon the formatting, I am a newbie yet to absorb all the formatting rules)
$z + a + b = d + e$
$a + b = 90 - 3z /2$
$d = 45^{circ} + z/4$
$e = 45^{circ} - 3z /4$
$f = 45^{circ} + 3z /4$
$i = 45^{circ} - z / 4$
$j = 67.5^{circ} + z / 8$
$l = 90^{circ} - z / 2 - b$
$x = 5z / 8 + b - 22.5^{circ} = 67.5^{circ} - 7z/8 -a$
I have verified that if I use the original Langley's apex angle $z = 20^{circ}$ and $a = 20^{circ}$ (or $b = 40^{circ}$) I get the correct answer as $x = 30^{circ}$.
So except for $l$ and $x$ all other angles are determined solely by the apex angle $z$. By construction it appears that these two should also be uniquely determined once the apex angle is fixed.
So my question is that:
Is there a way to derive angle $x$ ($BEF$) as a function angle $z$ ($BAC$) alone?
Alternately can angle $a$ or angle $b$ derived as a function of the apex angle $z$?
triangle angle
triangle angle
edited Jan 17 at 10:05
idriskameni
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
edited Jan 17 at 10:05
idriskameni
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
edited Jan 17 at 10:05
idriskameni
733319
edited Jan 17 at 10:05
idriskameni
733319
edited Jan 17 at 10:05
idriskameni
733319
733319
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
asked Jan 17 at 9:34
Jagadish ShriJagadish Shri
12
12
$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40
add a comment |
$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40
$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40
$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40
add a comment |
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$begingroup$
I will go through the links suggested. Thanks for the edit.
$endgroup$
– Jagadish Shri
Jan 17 at 10:40